A 105 L propane storage tank holds 3.8 kg of propane initially at a quality of 5

Question

A 105 L propane storage tank holds 3.8 kg of propane initially at a quality of 55%. The tank has a pressure relief valve at the top such that, if an internal pressure of 1.25 MPa is reached, the valve opens to release saturated vapor, and will open as needed to maintain the tank at a constant pressure of 1.25 MPa. The propane tank is left outside on a hot summer day, where the radiative heat transfer rate from the sun is 45 W. (a.) How much heat (in kJ) can be transferred to the propane before the valve opens? (b.) What is the temperature (in

Explanation / Answer

a) Given

V = 105 L and m = 3.8 kg

v1 = V/m = 0.105/3.8 = 0.02763 m^3/kg

Now the properties of propane at

x = 0.55 and v1 = 0.02763 m^3/kg are

P1 = 0.95 MPa

h1 = 150.4 kJ/kg

T1 = 24.92 deg

Saturation properties of Propane at P = 1.25 MPa

hf = -2.738 kJ/kg

hg = 311.7 kJ/kg

vf = 0.002109 m^3/kg

vg = 0.03649 m^3/kg

Tsat = 36.11 deg

The final state is Saturated Vapor condition

h2 = hg = 311.7 kJ/kg

Heat transferred = m*(h2-h1) = 3.8*(311.7-150.4) = 612.94 kJ

b) The final temperature will be Saturation temperation at P = 1.25 MPa

T2 = Tsat = 36.11 deg

c)Mass of liquid,mf = (1-x)*m = 0.45*3.8 = 1.71 kg

Heat required for mass of liquid will be,Q = mf*(h2-h1) = 1.71*(311.7-150.4)

Q = 272.823 kJ

Heat Supplied,Qs = P*t = 45*t

Qs = Q

45*t = 272823 J

t = 6062.73 s =1.684 hours

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